Crossing over and Linkage - TPR Science Workbook, Passage 33

[NA19]

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The black colour coat in hamsters is due to a dominant gene (B). A recessive allele (b) at this locus results in a brown coat when homozygous. However, neither coat colour is expressed when the organism is homozygous for the allele (a) at a separate locus. The a/a genotype results in a white (albino) coat, regardless of the allele at the B locus. The wile type (+) at the (a) locus allows normal coat coloration, whether the genotype is +/+ or +/a.

A female hamster with the genotype B/B; +/+ is crossed with a male hamster of genotype b/b; a/a. Female offspring from the F1 generation were backcrossed with b/b; a/a parent. The distribution of coat coloration among the progeny is as follows: black (66), brown (34), and white (100).

Based on these results, what is the genetic map distance (frequency of recombination) between the two loci discussed in this passage?

Ans: 34 centimorgans

I don't really understand crossing over/linkage q's so if anyone could help, that'd be great.

 

[ashtonjam]

When doing linkage problems, it's really helpful to write the chromosomes in phases with both alleles on the same chromosome. You know for sure that the parents are B+/B+ and ba/ba. When they mate, the offspring will be all B+/ba.

Re-crossing B+/ba offspring with their ba/ba parent can have different outcomes. Let's say that no crossing over happens. The two most numerous types of gametes from the B+/ba offspring are B+ and ba, which is pretty simple. When those gametes are combined with the parent's ba gamete (only one possibility), the genotypes are B+/ba and ba/ba. There's two possibilities.

Sometimes there is crossover. The number of crossover gametes will always be less than the number of non-crossover gametes. Crossing over B+/ba will give you two new gametes: Ba and b+. The crossover offspring will be Ba/ba and b+/ba.

Therefore, we've come up with 4 genotypes of offspring. Two of them are parental (non-crossover). Two of them are crossover (and less common).

Parental: B+/ba (Black) 66
Parental: ba/ba (Albino) 66
Crossover: Ba/ba (Albino) 34
Crossover: b+/ba (Brown) 34

Notice that the albinos actually have two different genotypes.

The trick is that the two parental genotypes will have the same probability of happening with each other. The same goes for crossover. I filled in the numbers next to the categories. Notice that you can go ahead and write the numbers for Black parental and Brown crossover right away. Then, you can assume that the number of parental Albinos is 66 (same as the other parental number) and the number of crossover Albinos is 34 (same as the other crossover number). It checks out since 66 + 34 is 100.

To determine centimorgans, you divide the number of crossover individuals by the total individuals.

68 / 200 = 34.

You actually could have shortcutted to the answer by realizing that brown individuals are only due to crossover. You could take 34 and double it to get the number of albino individuals, then divide by 200. But understanding it is probably the better initial choice.

 

[NA19]

How do you know what the crossover individuals are in general for these kinds of problems?

 

[Messier83]

When doing linkage problems, it's really helpful to write the chromosomes in phases with both alleles on the same chromosome. You know for sure that the parents are B+/B+ and ba/ba. When they mate, the offspring will be all B+/ba.

Re-crossing B+/ba offspring with their ba/ba parent can have different outcomes. Let's say that no crossing over happens. The two most numerous types of gametes from the B+/ba offspring are B+ and ba, which is pretty simple. When those gametes are combined with the parent's ba gamete (only one possibility), the genotypes are B+/ba and ba/ba. There's two possibilities.

Sometimes there is crossover. The number of crossover gametes will always be less than the number of non-crossover gametes. Crossing over B+/ba will give you two new gametes: Ba and b+. The crossover offspring will be Ba/ba and b+/ba.

Therefore, we've come up with 4 genotypes of offspring. Two of them are parental (non-crossover). Two of them are crossover (and less common).

Parental: B+/ba (Black) 66
Parental: ba/ba (Albino) 66
Crossover: Ba/ba (Albino) 34
Crossover: b+/ba (Brown) 34

Notice that the albinos actually have two different genotypes.

The trick is that the two parental genotypes will have the same probability of happening with each other. The same goes for crossover. I filled in the numbers next to the categories. Notice that you can go ahead and write the numbers for Black parental and Brown crossover right away. Then, you can assume that the number of parental Albinos is 66 (same as the other parental number) and the number of crossover Albinos is 34 (same as the other crossover number). It checks out since 66 + 34 is 100.

To determine centimorgans, you divide the number of crossover individuals by the total individuals.

68 / 200 = 34.

You actually could have shortcutted to the answer by realizing that brown individuals are only due to crossover. You could take 34 and double it to get the number of albino individuals, then divide by 200. But understanding it is probably the better initial choice.

Thoughtful explanation, thank you. But I had a further question. I thought that the probability (before crossing over) would be 1/4 Black, 1/4 brown and 1/2 albino. I arrived at these numbers using the standard dihybrid cross approach.

So for the F1 female, since they're all the same: Bb+a. The possible gametes are: B+, b+, Ba, and ba. The possible gametes for the parent we are backcrossing with are all ba. So again, when you fill in the punnett square for dihybrid crosses (the one with 16 slots), you end up with 1/4 of the offspring as brown because their genotypes are bb+a.

Perhaps you can see the flaw here?

 

[Messier83]

How do you know what the crossover individuals are in general for these kinds of problems?

Not sure if you are still studying for the MCAT, but to answer your question:

In general, the crossover individuals are the ones that you would otherwise not have expected to occur based on standard Mendelian genetics.

 

[DeathandTaxes]

Could someone explain to me how to tell what the crossover genotypes are?

Why is it that if there were no crossing over, no brown genotype would occur?

I'd prefer if you could explain it using the naming convention of the passage.

So we have female offspring from experiment backcrossed to the (b/b; a/a) parent.

So I set this up as B/b;+/a with b/b/; a/a

If you choose one allele from each, it should follow the regular mendelian dihybrid scenario. Why are there only two genotypes if no crossing over happens?